Now that a counterexample has been posted: might it still be true if $f$ is logarithmically convex? Equivalently (with $g=\log f\phantom.$): if $g: {\bf R}^n \rightarrow {\bf R}$ is convex, is the image of ${\rm grad}(g)$ a convex subset of ${\bf R}^n$?
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Noam D. ElkiesOct 27 '11 at 16:11

2 Answers
2

No. Consider $f(x,y)=e^x+y^2$, then $\varphi(x,y)=(e^x,2y)/(e^x+y^2)$. The image of $\varphi$ has only one point $(1,0)$ on the axis $y=0$. The points $a:=\varphi(0,1)=(\frac12,1)$ and $b:=\varphi(0,-1)=(\frac12,-1)$ belong to the image of $\varphi$ but their midpoint $\frac{a+b}2 = (\frac12,0)$ does not.

I have a similar problem. In my case, the function $f$ is
$$
f:\mathbb{R}^n \longrightarrow \mathbb{R}, f(X)=\sum_{i=1}^{m}a_i^2 \operatorname{e}^{2\langle X, \alpha_i \rangle}
$$
where $a_1, \ldots, a_m$ are not null, $\alpha_1, \dots, \alpha_m$ are any vectors in $\mathbb{R}^n$ and $\langle \cdot , \cdot \rangle$ is the usual inner product of $\mathbb{R}^n$.

I know that my problem is related to the Atiyah-Guillemin-Sternberg Convexity theorem, but I don't know how to prove it in my case by using elementary methods. I would really appreciate any comment.

Looks like this should be a new question, not an answer! Is the image of the logarithmic gradient here just the convex hull of $\lbrace 2 \alpha_i \rbrace$? If so then it can probably be proved using something like the Brouwer fixed-point theorem; I don't know how elementary that would be for your purposes.
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Noam D. ElkiesOct 28 '11 at 16:30